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Fixed Point Iteration

Extracts from this document...

Introduction

Fixed Point Iteration x = g(x)

I would like to solve the equation ½lnx + x - 10 = 0. The graph below shows the equation y = ½lnx + x – 10. As you can see there is only root, which is between [8, 9].

This can be rearranged to x = 10 – ½lnx. Let g(x) = 10 – ½lnx. In order to solve the original equation – ½lnx + x – 10 = 0 we can find the intersection of y = x and y = g(x)

The graph below showsy = g(x) and y=x.

By looking at the graph, I can see that the point of intersection is around the point when x = 8.

Let x1 = 8.

Middle

8.906604

8.906604

X9

8.906604

8.906604

X10

8.906604

8.906604

The rearrangement of ½lnx + x - 10 = 0 to x = 10 – ½lnx converged to the root 8.906604 because the gradient was not too steep in that region.

In order for the rearrangement to be successful we must have – 1 < g ‘ (x) < 1.

The gradient function of x = g(x) is                  g ‘ (x) = -0.5

In this case g ‘ (x) =      -0.5     _

8.906604

= - 0.0561 which is between [-1, 1]

The method found that x = 8.906604 to 7 significant figures.

In error bounds x = 8.9066035 ± 0.0000005

The method can also be shown graphically. This particular diagram is called a cobweb diagram.

The graphs below show the root more accurately.

Conclusion

ellspacing="0" class="c24">

x = g(x)

X1

8.9

9.0250135

X2

9.0250135

7.0284978

X3

7.0284978

381.07811

X4

381.07811

0

X5

0

485165195

X6

485165195

0

X7

0

485165195

X8

485165195

0

X9

0

485165195

This rearrangement fails to identify the root because the gradient was too steep at the point of intersection, which is obvious from the graph.

g(x) = e2(10-x) = e20-2x

The gradient function of g(x) is               g ‘ (x) = -2e20-2x

We need – 1 < g ‘ (x) < 1.

For example at the point x = 8.9,

g ‘ (x) = -2e20-(2x8.9)

= 9  which is very steep so the root is not identified.

As you can see from the diagram below the rearrangement fails to identify the root and does not converge towards it.

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